Consider the van der Waal's constants, a and b, for the following gases. Gas Ar Ne Kr Xe a / ( atmd m ^ 6 mo l ^ - 2 ) 1.3 0.2 5.1 4.1 b / ( 10 ^ - 2 d m ^ 3 mo l ^ - 1 ) 3.2 1.7 1.0 5.0 Which gas is expected to have

Correct answer: A. Kr Critical temperature is :- T _ C = (8a)/(27Rb) for Kr the value of ((a)/(b)) is the highest so, T _ c for Kr will be highest. (a)/(b) for Ar = (1.3)/(3.2) = 0.406 (a)/(b) for Ne = (0.2)/(1.7) = 0.118 (a)/(b) for Kr = (5.1)/(1) = 5.1 (a)/(b) for Xe = (4.1)/(5) = 0.82

JEE Main 20199 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 April
Shift: Morning Shift
Subject: Chemistry
Chapter: States of Matter

JEE Main 2019 · 9 April · Morning Shift · Chemistry · States of Matter

Consider the van der Waal's constants, a and b, for the following...

Consider the van der Waal's constants,

a

and

b,

for the following gases.

Gas	$Ar$	$Ne$	$Kr$	$Xe$
$a / ( atmd{m}^{6}mo{l}^{- 2} )$	$1.3$	$0.2$	$5.1$	$4.1$
$b / ( {10}^{- 2}d{m}^{3}mo{l}^{- 1} )$	$3.2$	$1.7$	$1.0$	$5.0$

Which gas is expected to have the highest critical temperature?

Options

A $Kr$
B $Ar$
C $Xe$
D $Ne$

Correct answer

A. $Kr$

Step-by-step solution

Critical temperature is :-

${T}_{C} = \dfrac{8a}{27Rb}$
for $Kr$ the value of $\left(\dfrac{a}{b}\right)$ is the highest so, ${T}_{c}$ for $Kr$ will be highest.
$\dfrac{a}{b}$ for $Ar = \dfrac{1.3}{3.2} = 0.406$
$\dfrac{a}{b}$ for $Ne = \dfrac{0.2}{1.7} = 0.118$
$\dfrac{a}{b}$ for $Kr = \dfrac{5.1}{1} = 5.1$
$\dfrac{a}{b}$ for $Xe = \dfrac{4.1}{5} = 0.82$

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